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NASA's solar sail successfully spreads its wings in space (www.space.com)

submitted 2 months ago by threelonmusketeers@sh.itjust.works to c/spaceflight@sh.itjust.works

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[–] nikaaa@lemmy.world 3 points 2 months ago

Oh yes, of course:

I guess the formula to calculate the momentum p of light is p = E/c where c is the speed of light and E is the energy.

So, if the Energy is 10 MW, or alternatively, 10 MJ/sec, then E = 1e7 J. c = 3e8 m/s. p = E/c = 1e7/3e8 J/(m/s) = 1e7•1e-17/3e8 kg•(m/s)²/(m/s) = 3e-2 kg•(m/s).

(Since E = m•c², 1 kg • c² = 1 kg • (3e8 m/s)² = 1e17 kg•(m/s)².)

Now, if we assume that that momentum gets transferred to the sail, which weighs approximately 20 kg, then we get the velocity of the sail: v = p/20 kg = 3e-2 kg•(m/s)/20 kg ≈ 1e-4 m/s. And this transfer of momentum happens every second, so the acceleration is roughly 1e-4 m/s², or 0.001 m/s².